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Monomials are indexed by ${\mathbb{N}}^n$, a countable set. A polynomial is a finite list of monomials with rational coefficients, so $P_n$ is a countable union of countable sets.

$X_1$ is the set of real algebraic numbers and is countable because each nonzero polynomial has finitely many roots. For $n\ge 2$, $X_n$ is uncountable: the polynomial $x_1$ vanishes on $\{0\}\times {\mathbb{R}}^{n-1}$.

Bilinearity and symmetry follow from the integral. Also $⟨p,p⟩={\int }_0^{1}p(t{)}^{2}dt\ge 0$, and equality forces the continuous polynomial $p$ to vanish identically.

Successive substitution shows every stage is a polynomial in $z=h\lambda$ times $y_n$, of degree at most its stage number. Hence $y_{n+1}=r(z)y_n$ with $r$ a degree-$\nu$ polynomial.

If $b^k=0$, then $(u+b{)}^{-1}=u^{-1}{\sum }_{j=0}^{k-1}(-b{u}^{-1}{)}^j$. For the polynomial, its positive-degree part is nilpotent by (ii), so adding it to the unit $a_0$ gives a unit.

One direction follows because finitely many nilpotent coefficients generate a nilpotent ideal. Conversely, reducing a nilpotent polynomial modulo $N$ gives a nilpotent polynomial over the reduced ring $R/N$; comparing its highest nonzero coefficient shows it must be zero, so every coefficient lay in $N$.

If $p=m_{{\alpha }_W}$ and $q=m_{{\alpha }_{V/W}}$, then $m_{\alpha }$ is divisible by both. Also $q(\alpha )V\subseteq W$, so $p(\alpha )q(\alpha )=0$; hence $\mathrm{lcm}(p,q)\mid m_{\alpha }\mid pq$. A single size-two nilpotent Jordan block with its invariant eigenline makes the lcm too small; a diagonal operator with the same eigenvalue on $W$ and $V/W$ makes the product too large. Square-free $m_{\alpha }$ implies both induced minimal polynomials square-free, so diagonalizability descends. The nilpotent-block example disproves the converse.

Generators define a surjection $R^k\to M$, and conversely images of the standard basis generate any quotient. Writing $\varphi (m_i)={\sum }_j{a}_{ij}{m}_j$ with $a_{ij}\in I$ and applying the adjugate to $XI-A$ gives a monic annihilating polynomial whose lower coefficients lie in $I$. Taking $\varphi =1$ when $IM=M$ yields $xM=0$ with $x-1\in I$; for $M=\mathbb{Z}/3$ and $I=(2)$ take $x=3$. The Jacobson radical criterion follows by placing a nonunit $1-rs$ in a maximal ideal. It yields Nakayama’s lemma. Applying the determinant trick to preimages of a finite generating set constructs a polynomial right inverse to any surjective endomorphism, proving injectivity.

Expanding the order condition about $w=1$ yields $\rho (w)={\sigma }_s{\sum }_{l=1}^s{w}^{s-l}(w-1{)}^l/l$ and ${\sigma }_s=(\sum 1/l{)}^{-1}$. Thus BDF2 has $\rho =w^2-4w/3+1/3$, ${\sigma }_2=2/3$, and BDF3 has $\rho =w^3-18w^2/11+9w/11-2/11$, ${\sigma }_3=6/11$. Their first characteristic polynomials satisfy the root condition, so consistency plus Dahlquist equivalence gives convergence. For BDF2 the stability boundary $z=\rho (e^{i\theta })/({\sigma }_2{e}^{2i\theta })$ has nonnegative real part; hence the whole left half-plane is stable.

Noetherian means every ideal is finitely generated, equivalently every ascending ideal chain stabilises. Quotients preserve this; for example $\mathbb{Z}[\sqrt{-5}]$ is Noetherian as a quotient of $\mathbb{Z}[X]$ but not a UFD. A UFD need not be Noetherian (a polynomial ring in infinitely many variables is a counterexample). Stabilisation of kernels proves every surjective endomorphism of a Noetherian ring injective. The shift $k[x_1,x_2,\dots ]\to k[x_1,x_2,\dots ]$, $x_1\mapsto 0$, $x_{i+1}\mapsto x_i$, is surjective noninjective; $x\mapsto x^2$ on $k[x]$ shows injective need not mean surjective. Finally $f_n(X)=\left(\genfrac{}{}{0ex}{}{X}{n}\right)$ is integer-valued. Finite differences prove uniquely that every integer-valued polynomial is an integral linear combination of the $f_n$, so they are a $\mathbb{Z}$-basis. Int$(\mathbb{Z})$ is not Noetherian: the denominators in $f_p$ yield an ideal chain requiring new prime denominators.

Put $\Gamma (z)=I(z)$ for $\Re z>0$. Integration by parts gives $\Gamma (z+1)=z\Gamma (z)$, both proving $\Gamma (n)=(n-1)!$ and defining the meromorphic continuation. A keyhole-contour evaluation of ${\int }_0^{\infty }{t}^{z-1}/(1+t)dt$ gives

Forthe reflection formula at $z$ and $z+\frac{1}{2}$ gives $g(z+1)-g(z)=-2\log 2$. Thus $h(z)=g(z)+2z\log 2$ is entire and $1$-periodic; poles and zeros have cancelled in the quotient. Its Fourier expansion and the exponential growth bound make it a finite Laurent polynomial in $e^{2\pi iz}$; the same growth bound for $e^h$ excludes every nonconstant mode. Hence $h=b$. At $z=1/2$, $e^b=2\sqrt{\pi }$, yielding the duplication formula.

Set $z=1/6$ and use reflection at $1/3$:Therefore

Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.

For a same-degree counterexample, let $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$,Both split in $L$, but the first action has two orbits of size two while the second is transitive, so the subgroups are not conjugate in $S_4$.

Over a finite field, Frobenius cycle type proves the assertion. The given quartic reduces over ${\mathbb{F}}_2$ to $X^4+X^3+1$, which is irreducible, so its group there is cyclic of order four generated by a $4$-cycle.

Over $\mathbb{Q}$ the polynomial is irreducible by the same reduction. Modulo $3$ it factors as linear times irreducible cubic, so Dedekind factorization cycle test gives a $3$-cycle; modulo $2$ it gives a $4$-cycle. Its discriminant is $229$, and modulo $23$ its factor degrees are $1,1,2$, giving a transposition. These cycle types force the transitive group to be $S_4$.

The fixed points are $(0,0)$ and $(\pm |\mu |,{\mu }^2)$. At the origin the eigenvalues are ${\mu }^2$ and $-1$, so it is a saddle. At either nonzero point the characteristic polynomial isThey are stable nodes for $0<|\mu |<1/(2\sqrt{2})$, degenerate nodes at equality, and stable spirals above it.

The stable manifold of the origin is exactly $x=0$. Write the stable manifold in the unstable direction as $y=a{x}^2+b{x}^4+O(x^6)$. Graph invariance givesThus $W^u$ has

For odd $n>1$, the primitive $2n$th roots are exactly the negatives of primitive $n$th roots. Since $\phi (2n)=\phi (n)$, their monic root polynomials satisfy ${\Phi }_{2n}(X)={\Phi }_n(-X)$.

Runge theorem says that if $K\subset \mathbb{C}$ is compact and $\mathbb{C}\setminus K$ is connected, every function holomorphic near $K$ is uniformly approximable on $K$ by polynomials.

Choose a compact exhaustion $K_j$ of the slit disc $\Delta \setminus L^{+}$, rounding the two sides of the slit so that $0\notin K_j$ and $\mathbb{C}\setminus K_j$ is connected. Runge applied to $1/z$ gives a polynomial $p_j$ with ${\sup }_{K_j}|p_j-1/z|<1/j$. Every compact subset of the slit disc lies in all sufficiently large $K_j$, proving locally uniform convergence.